Let
$$
A_1\twoheadrightarrow
A_2\twoheadrightarrow
A_3\twoheadrightarrow
A_4\twoheadrightarrow
\cdots
$$
be an inductive sequence of countable abelian groups, the connecting homomorphisms of which are surjective and split, that is, we have embeddings $A_{n+1}\rightarrowtail A_n$ such that the composition $A_{n+1}\rightarrowtail A_n\twoheadrightarrow A_{n+1}$ is the identity for every $n$. This means that $A_{n+1}$ is a direct summand of $A_n$.

Let $\varinjlim A_n$ denote the inductive limit of the system
$$
A_1\twoheadrightarrow
A_2\twoheadrightarrow
A_3\twoheadrightarrow
A_4\twoheadrightarrow
\cdots
$$
and let $\varprojlim A_n$ denote the projective limit of the system
$$
A_1\leftarrowtail
A_2\leftarrowtail
A_3\leftarrowtail
A_4\leftarrowtail
\cdots.
$$
We get an induced map
$$
\varprojlim A_n\to\varinjlim A_n.
$$
As Zhen Lin has shown in over here, this map need not be surjective. Here is a weaker question:

Question:
If we have $\varinjlim A_n=0$, then can we conclude that $\varprojlim A_n=0$?

This would, of course, follow if the map $\varprojlim A_n\to\varinjlim A_n$ was always injective. Is there any reason to expect this?

is a sequence of inclusions of nested subgroups, so $\varprojlim A_n$ is just the intersection. An element of the kernel of $\varprojlim A_n\to\varinjlim A_n$ is just an element $a$ of $\bigcap A_n$ that is in the kernel of the map $A_1\twoheadrightarrow A_k$ for some $k$. But this implies $a=0$ since this map is a splitting of the inclusion $A_k\rightarrowtail A_1$.